Câu 5:

\(VT=\dfrac{x^2yz}{xy+x^2yz+xyz}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{xz+z+1}\\ =\dfrac{xz}{1+z+xz}+\dfrac{1}{z+1+xz}+\dfrac{z}{zx+z+1}\\ =\dfrac{zx+z+1}{zx+z+1}=1\)

1 watched

2 locking - was - dropped

4 would finish

5 write - will try - to make

5 left

6 Does - asspoint

III

1 handle

2 satisfying

3 conservation

Nhỏ quá

Bài 1: 

a: \(=y\left(4x^2-1\right)=y\left(2x-1\right)\left(2x+1\right)\)

2:

a: BC=căn 15^2+20^2=25cm

AH=15*20/25=12cm

góc ADH=góc AEH=góc DAE=90 độ

=>ADHE là hình chữ nhật

=>DE=AH=12cm

b: ΔAHB vuông tại H có HD vuông góc AB

nên AD*AB=AH^2

ΔAHC vuông tại H có HE vuông góc AC

nên AE*AC=AH^2

=>AD*AB=AE*AC

c: góc IAC+góc AED

=góc ICA+góc AHD

=góc ACB+góc ABC=90 độ

=>AI vuông góc ED

4:

a: góc BDH=góc BEH=góc DBE=90 độ

=>BDHE là hình chữ nhật

b: BDHE là hình chữ nhật

=>góc BED=góc BHD=góc A

Xét ΔBED và ΔBAC có 

góc BED=góc A

góc EBD chung

=>ΔBED đồng dạng với ΔBAC
=>BE/BA=BD/BC

=>BE*BC=BA*BD

c: góc MBC+góc BED

=góc C+góc BHD

=góc C+góc A=90 độ

=>BM vuông góc ED

P6:

1. big

2. been

3. bought

4. people

P7:

1. T

2. F

3. F

4. F

ĐKXĐ: \(-1\le x\le3\)

Đặt \(\sqrt{x+1}+\sqrt{3-x}=t\ge\sqrt{x+1+3-x}=2\)

\(\Rightarrow4+2\sqrt{-x^2+2x+3}=t^2\)

\(\Rightarrow\sqrt{-x^2+2x+3}=\dfrac{t^2-4}{2}\) (1)

Phương trình trở thành:

\(t-\dfrac{t^2-4}{2}=2\)

\(\Leftrightarrow2t-t^2=0\Rightarrow\left[{}\begin{matrix}t=0\left(loại\right)\\t=2\end{matrix}\right.\)

Thế vào (1):

\(\Rightarrow\sqrt{-x^2+2x+3}=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)

Bài 1: 

a) Ta có: \(M=\left(\dfrac{x+2}{x^2+2x+1}+\dfrac{x-2}{1-x^2}\right)\cdot\dfrac{x+1}{x}\)

\(=\left(\dfrac{\left(x+2\right)\left(x-1\right)}{\left(x+1\right)^2\cdot\left(x-1\right)}-\dfrac{\left(x-2\right)\left(x+1\right)}{\left(x+1\right)^2\cdot\left(x-1\right)}\right)\cdot\dfrac{x+1}{x}\)

\(=\dfrac{x^2-x+2x-2-\left(x^2+x-2x-2\right)}{\left(x+1\right)^2\cdot\left(x-1\right)}\cdot\dfrac{x+1}{x}\)

\(=\dfrac{x^2+x-2-x^2+x+2}{\left(x+1\right)\left(x-1\right)}\cdot\dfrac{1}{x}\)

\(=\dfrac{2x}{\left(x+1\right)\left(x-1\right)}\cdot\dfrac{1}{x}\)

\(=\dfrac{2}{x^2-1}\)

Bài 2:

1: Ta có: \(\left(x-5\right)^2+\left(x+3\right)^2=2\left(x-4\right)\left(x+4\right)-5x+7\)

\(\Leftrightarrow x^2-10x+25+x^2+6x+9=2\left(x^2-16\right)-5x+7\)

\(\Leftrightarrow2x^2-4x+34=2x^2-32-5x+7\)

\(\Leftrightarrow2x^2-4x+34-2x^2+5x+25=0\)

\(\Leftrightarrow x+59=0\)

hay x=-59

Vậy: S={-59}

a) -28xy⁵z³ : 7xy²z³= -4y³

b) 8x²y²z : 6xyz= 6xy

c) 6x³y⁴ : x³y=6y³

d) 30x²y²z: 6xyz

e) 54x⁴y²z: 9x⁴y= 6y³z

f) x⁴y³z² : 3xyz²= \(\dfrac{1}{2}\)x³y²

g) (x+y)⁵: (x+y)⁴= (x+y)¹= x+y

h) (x-y)⁷ : (y-x)⁶ = -(y-x)⁷: (y-x)⁶= -(y-x)¹= -(y-x) = -y +x

i) (x-y+z)⁵ : (x-y+z)⁴ = (x-y+z)¹= (x+y-z)= x+y-z

ĐK: \(\left\{{}\begin{matrix}x\ne-y\\y\ge\dfrac{3}{2}\end{matrix}\right.\).

\(\left\{{}\begin{matrix}\dfrac{2x-y+3}{x+y}=1\\2x-\sqrt{2y-3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2x-y+3}{x+y}-1=0\\2x-\sqrt{2y-3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2x-y+3}{x+y}-\dfrac{x+y}{x+y}=0\\2x-\sqrt{2y-3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y+3-x-y=0\\2x-\sqrt{2y-3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2y+3=0\\2x-\sqrt{2y-3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-\left(2y-3\right)=0\\2x-\sqrt{2y-3}=0\end{matrix}\right..\)

Đặt a = x, b = \(\sqrt{2y-3}\).

Hệ phương trình trở thành: \(\left\{{}\begin{matrix}a-b^2=0\\2a-b=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b^2\\2b^2-b=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b^2\\b\left(2b-1\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b^2\\\left[{}\begin{matrix}b=0\\b=\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\left\{{}\begin{matrix}\left[{}\begin{matrix}a=0\\a=\dfrac{1}{4}\end{matrix}\right.\\\left[{}\begin{matrix}b=0\\b=\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\\\left[{}\begin{matrix}y=\dfrac{3}{2}\\2y-3=\dfrac{1}{4}\end{matrix}\right.\end{matrix}\right.\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\\\left[{}\begin{matrix}y=\dfrac{3}{2}\\2y=\dfrac{13}{4}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\\\left[{}\begin{matrix}y=\dfrac{3}{2}\\y=\dfrac{13}{8}\end{matrix}\right.\end{matrix}\right..\)

Vậy hệ phương trình có nghiệm (x;y) \(\in\) \(\left\{\left(0;\dfrac{3}{2}\right),\left(\dfrac{1}{4};\dfrac{13}{8}\right)\right\}\).